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"solution": "More generally,\n\\[\\begin{pmatrix} 1 & a \\\\ 0 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & b \\\\ 0 & 1 \\end{pmatrix} = \\begin{pmatrix} 1 & a + b \\\\ 0 & 1 \\end{pmatrix}.\\]Therefore,\n\\[\\begin{pmatrix} 1 & 1 \\\\ 0 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & 3 \\\\ 0 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & 5 \\\\ 0 & 1 \\end{pmatrix} \\dotsm \\begin{pmatrix} 1 & 99 \\\\ 0 & 1 \\end{pmatrix} = \\begin{pmatrix} 1 & 1 + 3 + 5 + \\dots + 99 \\\\ 0 & 1 \\end{pmatrix} = \\boxed{\\begin{pmatrix} 1 & 2500 \\\\ 0 & 1 \\end{pmatrix}}.\\]" |